Is ${145344}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {145344}= &&{1}\cdot100000+ \\&&{4}\cdot10000+ \\&&{5}\cdot1000+ \\&&{3}\cdot100+ \\&&{4}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {145344}= &&{1}(99999+1)+ \\&&{4}(9999+1)+ \\&&{5}(999+1)+ \\&&{3}(99+1)+ \\&&{4}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {145344}= &&\gray{1\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {1}+{4}+{5}+{3}+{4}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${145344}$ is divisible by $3$ if ${ 1}+{4}+{5}+{3}+{4}+{4}$ is divisible by $3$ Add the digits of ${145344}$ $ {1}+{4}+{5}+{3}+{4}+{4} = {21} $ If ${21}$ is divisible by $3$ , then ${145344}$ must also be divisible by $3$ ${21}$ is divisible by $3$, therefore ${145344}$ must also be divisible by $3$.